Bounded variation and satisfying intermediate value theorem, then continuous If a real-valued function F defined on a closed interval [a,b] is of bounded variation and satisfies the intermediate value theorem, then is it true that F is continuous?
It seems intuitively clear, but I cannot prove it..
 A: Try turning it around. Suppose that $f$ has the intermediate value property but is discontinuous at some point $x_0 \in [a,b]$. For simplicity take $x_0 = a$. Then there exist $\epsilon > 0$ and a sequence $x_k \searrow a$ with the property that $|f(x_k) - f(a)| > \epsilon$ for all $k$.
Define a partition of $[a,x_1]$ as follows. Let $n \ge 2$ be even, set $t_0 = a$, and $t_n = x_1$.
Since $|f(t_n) - f(t_0)| > \epsilon$, there exists a point $t_{n-1} \in (t_0,t_n)$ with the property that $|f(t_{n-1}) - f(t_0)| < \frac{\epsilon}2$. Thus $$|f(t_{n-1}) - f(t_n)| > \frac{\epsilon}{2}.$$
Next select $t_{n-2} \in (t_0,t_{n-1})$ satisfying $|f(t_{n-2}) - f(t_0)| > \epsilon$ (use one of the $x_k$'s in the sequence) and proceed as above to find a point $t_{n-3} \in (t_0,t_{n-2})$ satisfying $$|f(t_{n-3}) - f(t_{n-2})| > \frac{\epsilon}{2}.$$
Proceed inductively until you have defined points $t_0,t_1,\ldots t_n$. At each step you alternate between using the IV property and discontinuity to get a zig-zag of function values. Moreover $$\sum_{k=1}^n |f(t_{k-1}) - f(t_k)| \ge |f(t_1) - f(t_2)| + |f(t_3) - f(t_4)| + \cdots + |f(t_{n-1}) - f(t_n)| > n \frac{\epsilon}{4}.$$
Thus $$V_{[a,b]} f \ge V_{[a,x_1]} f > n \frac{\epsilon}4$$ for all $n$ so that $V_{[a,b]} f = \infty$.
